How many grams of lithium bromide are present in a 15.50 mole sample? <\/p>\n\n\n\n
2. How many moles of copper (II) cyanide are present in a 3.40 x 10 cubed mg sample? <\/p>\n\n\n\n
3. How many moles of cobalt (1) carbonate are present in a sample containing 1.75 g of cobalt?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s break down and solve each of these questions step-by-step.<\/p>\n\n\n\n We are asked to find the mass of lithium bromide (LiBr) in grams.<\/p>\n\n\n\n Step 1: Use the molar mass of LiBr.<\/strong><\/p>\n\n\n\n The molar mass of lithium bromide is calculated by adding the atomic masses of lithium (Li) and bromine (Br):<\/p>\n\n\n\n So, the molar mass of LiBr = 6.94 g\/mol + 79.90 g\/mol = 86.84 g\/mol<\/strong>.<\/p>\n\n\n\n Step 2: Apply the formula:<\/strong> [ \\text{Mass} = 15.50 \\, \\text{mol} \\times 86.84 \\, \\text{g\/mol} ]<\/p>\n\n\n\n [ \\text{Mass} = 1346.04 \\, \\text{g} ]<\/p>\n\n\n\n Answer<\/strong>: There are 1346.04 grams<\/strong> of lithium bromide in a 15.50 mole sample.<\/p>\n\n\n\n We are asked to find the number of moles of copper (II) cyanide (Cu(CN)\u2082) in a given mass of 3.40 x 10\u00b3 milligrams (mg).<\/p>\n\n\n\n Step 1: Convert milligrams to grams.<\/strong><\/p>\n\n\n\n 1 gram = 1000 milligrams, so: Step 2: Find the molar mass of copper (II) cyanide.<\/strong><\/p>\n\n\n\n The molar mass of Cu(CN)\u2082 is calculated by adding the atomic masses of copper (Cu), carbon (C), and nitrogen (N):<\/p>\n\n\n\n So, the molar mass of Cu(CN)\u2082: Step 3: Use the formula to find moles.<\/strong> Answer<\/strong>: There are 0.0294 moles<\/strong> of copper (II) cyanide in the 3.40 x 10\u00b3 mg sample.<\/p>\n\n\n\n We are asked to find the number of moles of cobalt (I) carbonate (Co\u2082CO\u2083) from the given cobalt mass.<\/p>\n\n\n\n Step 1: Find the molar mass of cobalt (I) carbonate.<\/strong><\/p>\n\n\n\n Cobalt (I) carbonate has the formula Co\u2082CO\u2083, where:<\/p>\n\n\n\n The molar mass of Co\u2082CO\u2083: Step 2: Use the formula to find moles of cobalt.<\/strong><\/p>\n\n\n\n We know that cobalt (I) carbonate contains two moles of cobalt per formula unit. First, we need to find the moles of cobalt (Co) in the sample.<\/p>\n\n\n\n [ Step 3: Find the moles of cobalt (I) carbonate.<\/strong><\/p>\n\n\n\n Since each mole of Co\u2082CO\u2083 contains 2 moles of cobalt, the number of moles of cobalt (I) carbonate is: Answer<\/strong>: There are 0.01485 moles<\/strong> of cobalt (I) carbonate in the sample containing 1.75 g of cobalt.<\/p>\n\n\n\n How many grams of lithium bromide are present in a 15.50 mole sample? 2. How many moles of copper (II) cyanide are present in a 3.40 x 10 cubed mg sample? 3. How many moles of cobalt (1) carbonate are present in a sample containing 1.75 g of cobalt? The Correct Answer and Explanation is […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-186297","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186297","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=186297"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186297\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186297"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186297"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186297"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Question 1: How many grams of lithium bromide are present in a 15.50 mole sample?<\/h3>\n\n\n\n
\n
[ \\text{Mass (g)} = \\text{Moles (mol)} \\times \\text{Molar Mass (g\/mol)} ]<\/p>\n\n\n\n
\n\n\n\nQuestion 2: How many moles of copper (II) cyanide are present in a 3.40 x 10\u00b3 mg sample?<\/h3>\n\n\n\n
[ 3.40 \\times 10^3 \\, \\text{mg} = \\frac{3.40 \\times 10^3}{1000} = 3.40 \\, \\text{g} ]<\/p>\n\n\n\n\n
[ 63.55 \\, \\text{g\/mol} + (2 \\times (12.01 \\, \\text{g\/mol} + 14.01 \\, \\text{g\/mol})) ]
[ = 63.55 \\, \\text{g\/mol} + (2 \\times 26.02 \\, \\text{g\/mol}) ]
[ = 63.55 \\, \\text{g\/mol} + 52.04 \\, \\text{g\/mol} ]
[ = 115.59 \\, \\text{g\/mol} ]<\/p>\n\n\n\n
[
\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar Mass (g\/mol)}}
]
[
\\text{Moles} = \\frac{3.40 \\, \\text{g}}{115.59 \\, \\text{g\/mol}} \\approx 0.0294 \\, \\text{mol}
]<\/p>\n\n\n\n
\n\n\n\nQuestion 3: How many moles of cobalt (I) carbonate are present in a sample containing 1.75 g of cobalt?<\/h3>\n\n\n\n
\n
[
(2 \\times 58.93 \\, \\text{g\/mol}) + 12.01 \\, \\text{g\/mol} + (3 \\times 16.00 \\, \\text{g\/mol})
]
[
= 117.86 \\, \\text{g\/mol} + 12.01 \\, \\text{g\/mol} + 48.00 \\, \\text{g\/mol}
]
[
= 177.87 \\, \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{Moles of Co} = \\frac{\\text{Mass of Co (g)}}{\\text{Molar Mass of Co (g\/mol)}}
]
[
\\text{Moles of Co} = \\frac{1.75 \\, \\text{g}}{58.93 \\, \\text{g\/mol}} \\approx 0.0297 \\, \\text{mol}
]<\/p>\n\n\n\n
[
\\text{Moles of Co\u2082CO\u2083} = \\frac{\\text{Moles of Co}}{2} = \\frac{0.0297 \\, \\text{mol}}{2} \\approx 0.01485 \\, \\text{mol}
]<\/p>\n\n\n\n
\n\n\n\nSummary:<\/h3>\n\n\n\n
\n